109 views

Is the given statement True ?  Please explain

• $\forall_x \left \{ P(x) \vee Q(x) \right \}\Leftrightarrow \forall_x P(x) \vee \forall_x Q(x)$
• $\forall_x \left \{ P(x) \wedge Q(x) \right \}\Leftrightarrow \forall_x P(x) \wedge \forall_x Q(x)$
• $\exists_x \left \{ P(x) \vee Q(x)\right \} \Leftrightarrow \exists_x P(x) \vee \forall_x Q(x)$
• $\exists_x \left \{ P(x) \wedge Q(x) \right \}\Leftrightarrow \exists_x P(x) \wedge \exists_x Q(x)$

Only 2nd and third is true
∀ is true when it is true for each nd every value of its domain so it is distributive on AND
∃ is true when it is true for atleast value of its domain  so it is distributive on OR
answered by Veteran (11.8k points) 13 42 139
selected
Can u tell why 4 is true ? with some reason / example .? Im not able to understsand that situation
bt 4 is nt true
∃x p(x)∧∃x q(x) and  ∃x (p(x)∧q(x))
suppose domain consist of all positive integers
p(x)= x is odd   q(x)= x is even
consider ∃x p(x)∧∃x q(x) ,it is saying that there exist odd numbers and there exist even numbers
now consider ∃x (p(x)∧q(x)) it is saying that there exist a no. which is is both odd and even it is trivially false.
@Dq.

P: Divisible by 2

Q: Divisible by 3

And let x={2,4,8}

For this 4th statment is   F$\Leftrightarrow$ T
Got it :) Thanks.
Can I ask one more question ?
yes bro

How to solve  these kind of questions ?

Some Resources I have seen some of the options is proved by CP or IP  rule and some options by examples . I m not getting a general idea how to deal with these kinda questionz :(

Either try with examples or you need to prove it(which is difficult)

Its hard for me try with examples also :(
Is there any other approach ?

Can u just show for this http://gateoverflow.in/3783/gate2005-it-36

How to check other options using examples ? :(
Sorry i was asking too much questions . I was not able to understand this topic at all

They have given example for 2005 question. Try it on your own. You will get it
Even I find difficult taking examples.